3.1.60 \(\int \frac {1}{(a+b \cos (c+d x))^{2/3}} \, dx\) [60]

Optimal. Leaf size=105 \[ \frac {\sqrt {2} F_1\left (\frac {1}{2};\frac {1}{2},\frac {2}{3};\frac {3}{2};\frac {1}{2} (1-\cos (c+d x)),\frac {b (1-\cos (c+d x))}{a+b}\right ) \left (\frac {a+b \cos (c+d x)}{a+b}\right )^{2/3} \sin (c+d x)}{d \sqrt {1+\cos (c+d x)} (a+b \cos (c+d x))^{2/3}} \]

[Out]

AppellF1(1/2,2/3,1/2,3/2,b*(1-cos(d*x+c))/(a+b),1/2-1/2*cos(d*x+c))*((a+b*cos(d*x+c))/(a+b))^(2/3)*sin(d*x+c)*
2^(1/2)/d/(a+b*cos(d*x+c))^(2/3)/(1+cos(d*x+c))^(1/2)

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Rubi [A]
time = 0.05, antiderivative size = 105, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.214, Rules used = {2744, 144, 143} \begin {gather*} \frac {\sqrt {2} \sin (c+d x) \left (\frac {a+b \cos (c+d x)}{a+b}\right )^{2/3} F_1\left (\frac {1}{2};\frac {1}{2},\frac {2}{3};\frac {3}{2};\frac {1}{2} (1-\cos (c+d x)),\frac {b (1-\cos (c+d x))}{a+b}\right )}{d \sqrt {\cos (c+d x)+1} (a+b \cos (c+d x))^{2/3}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + b*Cos[c + d*x])^(-2/3),x]

[Out]

(Sqrt[2]*AppellF1[1/2, 1/2, 2/3, 3/2, (1 - Cos[c + d*x])/2, (b*(1 - Cos[c + d*x]))/(a + b)]*((a + b*Cos[c + d*
x])/(a + b))^(2/3)*Sin[c + d*x])/(d*Sqrt[1 + Cos[c + d*x]]*(a + b*Cos[c + d*x])^(2/3))

Rule 143

Int[((a_) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_), x_Symbol] :> Simp[((a + b*x)
^(m + 1)/(b*(m + 1)*(b/(b*c - a*d))^n*(b/(b*e - a*f))^p))*AppellF1[m + 1, -n, -p, m + 2, (-d)*((a + b*x)/(b*c
- a*d)), (-f)*((a + b*x)/(b*e - a*f))], x] /; FreeQ[{a, b, c, d, e, f, m, n, p}, x] &&  !IntegerQ[m] &&  !Inte
gerQ[n] &&  !IntegerQ[p] && GtQ[b/(b*c - a*d), 0] && GtQ[b/(b*e - a*f), 0] &&  !(GtQ[d/(d*a - c*b), 0] && GtQ[
d/(d*e - c*f), 0] && SimplerQ[c + d*x, a + b*x]) &&  !(GtQ[f/(f*a - e*b), 0] && GtQ[f/(f*c - e*d), 0] && Simpl
erQ[e + f*x, a + b*x])

Rule 144

Int[((a_) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_), x_Symbol] :> Dist[(e + f*x)^
FracPart[p]/((b/(b*e - a*f))^IntPart[p]*(b*((e + f*x)/(b*e - a*f)))^FracPart[p]), Int[(a + b*x)^m*(c + d*x)^n*
(b*(e/(b*e - a*f)) + b*f*(x/(b*e - a*f)))^p, x], x] /; FreeQ[{a, b, c, d, e, f, m, n, p}, x] &&  !IntegerQ[m]
&&  !IntegerQ[n] &&  !IntegerQ[p] && GtQ[b/(b*c - a*d), 0] &&  !GtQ[b/(b*e - a*f), 0]

Rule 2744

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Dist[Cos[c + d*x]/(d*Sqrt[1 + Sin[c + d*x]]*Sqrt
[1 - Sin[c + d*x]]), Subst[Int[(a + b*x)^n/(Sqrt[1 + x]*Sqrt[1 - x]), x], x, Sin[c + d*x]], x] /; FreeQ[{a, b,
c, d, n}, x] && NeQ[a^2 - b^2, 0] &&  !IntegerQ[2*n]

Rubi steps

\begin {align*} \int \frac {1}{(a+b \cos (c+d x))^{2/3}} \, dx &=-\frac {\sin (c+d x) \text {Subst}\left (\int \frac {1}{\sqrt {1-x} \sqrt {1+x} (a+b x)^{2/3}} \, dx,x,\cos (c+d x)\right )}{d \sqrt {1-\cos (c+d x)} \sqrt {1+\cos (c+d x)}}\\ &=-\frac {\left (\left (-\frac {a+b \cos (c+d x)}{-a-b}\right )^{2/3} \sin (c+d x)\right ) \text {Subst}\left (\int \frac {1}{\sqrt {1-x} \sqrt {1+x} \left (-\frac {a}{-a-b}-\frac {b x}{-a-b}\right )^{2/3}} \, dx,x,\cos (c+d x)\right )}{d \sqrt {1-\cos (c+d x)} \sqrt {1+\cos (c+d x)} (a+b \cos (c+d x))^{2/3}}\\ &=\frac {\sqrt {2} F_1\left (\frac {1}{2};\frac {1}{2},\frac {2}{3};\frac {3}{2};\frac {1}{2} (1-\cos (c+d x)),\frac {b (1-\cos (c+d x))}{a+b}\right ) \left (\frac {a+b \cos (c+d x)}{a+b}\right )^{2/3} \sin (c+d x)}{d \sqrt {1+\cos (c+d x)} (a+b \cos (c+d x))^{2/3}}\\ \end {align*}

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Mathematica [A]
time = 0.12, size = 116, normalized size = 1.10 \begin {gather*} -\frac {3 F_1\left (\frac {1}{3};\frac {1}{2},\frac {1}{2};\frac {4}{3};\frac {a+b \cos (c+d x)}{a-b},\frac {a+b \cos (c+d x)}{a+b}\right ) \sqrt {-\frac {b (-1+\cos (c+d x))}{a+b}} \sqrt {\frac {b (1+\cos (c+d x))}{-a+b}} \sqrt [3]{a+b \cos (c+d x)} \csc (c+d x)}{b d} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a + b*Cos[c + d*x])^(-2/3),x]

[Out]

(-3*AppellF1[1/3, 1/2, 1/2, 4/3, (a + b*Cos[c + d*x])/(a - b), (a + b*Cos[c + d*x])/(a + b)]*Sqrt[-((b*(-1 + C
os[c + d*x]))/(a + b))]*Sqrt[(b*(1 + Cos[c + d*x]))/(-a + b)]*(a + b*Cos[c + d*x])^(1/3)*Csc[c + d*x])/(b*d)

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Maple [F]
time = 0.03, size = 0, normalized size = 0.00 \[\int \frac {1}{\left (a +b \cos \left (d x +c \right )\right )^{\frac {2}{3}}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a+b*cos(d*x+c))^(2/3),x)

[Out]

int(1/(a+b*cos(d*x+c))^(2/3),x)

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*cos(d*x+c))^(2/3),x, algorithm="maxima")

[Out]

integrate((b*cos(d*x + c) + a)^(-2/3), x)

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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*cos(d*x+c))^(2/3),x, algorithm="fricas")

[Out]

integral((b*cos(d*x + c) + a)^(-2/3), x)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{\left (a + b \cos {\left (c + d x \right )}\right )^{\frac {2}{3}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*cos(d*x+c))**(2/3),x)

[Out]

Integral((a + b*cos(c + d*x))**(-2/3), x)

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*cos(d*x+c))^(2/3),x, algorithm="giac")

[Out]

integrate((b*cos(d*x + c) + a)^(-2/3), x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {1}{{\left (a+b\,\cos \left (c+d\,x\right )\right )}^{2/3}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a + b*cos(c + d*x))^(2/3),x)

[Out]

int(1/(a + b*cos(c + d*x))^(2/3), x)

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